7.4 Regular Singular Points: Euler Equations

This section sets the stage for Sections 1.5, 1.6, and 1.7. If you’re not interested in those sections, but wish to learn about Euler equations, omit the introductory paragraphs and start reading at Definition 7.4.2.

In the next three sections we’ll continue to study equations of the form

\begin{equation} P_0(x)y''+P_1(x)y'+P_2(x)y=0 \tag{7.4.1}\end{equation}

where $P_0$, $P_1$, and $P_2$ are polynomials, but the emphasis will be different from that of Sections 7.2 and 7.3, where we obtained solutions of (7.4.1) near an ordinary point $x_0$ in the form of power series in $x-x_0$. If $x_0$ is a singular point of (7.4.1) (that is, if $P(x_0)=0$), the solutions can’t in general be represented by power series in $x-x_0$. Nevertheless, it’s often necessary in physical applications to study the behavior of solutions of (7.4.1) near a singular point. Although this can be difficult in the absence of some sort of assumption on the nature of the singular point, equations that satisfy the requirements of the next definition can be solved by series methods discussed in the next three sections. Fortunately, many equations arising in applications satisfy these requirements.

Definition 7.4.1

Let $P_0$, $P_1$, and $P_2$ be polynomials with no common factor and suppose $P_0(x_0)=0$. Then $x_0$ is a regular singular point of the equation

\begin{equation} P_0(x)y''+P_1(x)y'+P_2(x)y=0 \tag{7.4.2}\end{equation}

if $\eqref{eq:7.4.2}$ can be written as

\begin{equation} (x-x_0)^2A(x)y''+(x-x_0)B(x)y'+C(x)y=0 \tag{7.4.3}\end{equation}

where $A$, $B$, and $C$ are polynomials and $A(x_0)\ne0$; otherwise, $x_0$ is an irregular singular point of (7.4.2).

Example 7.4.1

Bessel’s equation,

\begin{equation} x^2y''+xy'+(x^2-\nu^2)y=0, \tag{7.4.4}\end{equation}

has the singular point $x_0=0$. Since this equation is of the form (7.4.3) with $x_0=0$, $A(x)=1$, $B(x)=1$, and $C(x)=x^2-\nu^2$, it follows that $x_0=0$ is a regular singular point of (7.4.4).

Example 7.4.2

Legendre’s equation,

\begin{equation} (1-x^2)y''-2xy'+\alpha(\alpha+1)y=0, \tag{7.4.5}\end{equation}

has the singular points $x_0=\pm1$. Mutiplying through by $1-x$ yields

\[ (x-1)^2(x+1)y''+2x(x-1)y'-\alpha(\alpha+1)(x-1)y=0, \]

which is of the form (7.4.3) with $x_0=1$, $A(x)=x+1$, $B(x)=2x$, and $C(x)=-\alpha(\alpha+1)(x-1)$. Therefore $x_0=1$ is a regular singular point of (7.4.5). We leave it to you to show that $x_0=-1$ is also a regular singular point of (7.4.5).

Example 7.4.3

The equation

\[ x^3y''+xy'+y=0 \]

has an irregular singular point at $x_0=0$. (Verify.)

For convenience we restrict our attention to the case where $x_0=0$ is a regular singular point of (7.4.2). This isn’t really a restriction, since if $x_0\ne0$ is a regular singular point of (7.4.2) then introducing the new independent variable $t=x-x_0$ and the new unknown $Y(t)=y(t+x_0)$ leads to a differential equation with polynomial coefficients that has a regular singular point at $t_0=0$. This is illustrated in Exercise 22 for Legendre’s equation, and in Exercise 23 for the general case.

Euler Equations

The simplest kind of equation with a regular singular point at $x_0=0$ is the Euler equation, defined as follows.

Definition 7.4.2

An Euler equation is an equation that can be written in the form

\begin{equation} ax^2y''+bxy'+cy=0, \tag{7.4.6}\end{equation}

where $a,b$, and $c$ are real constants and $a\ne0$.

Theorem 5.1.1 implies that (7.4.6) has solutions defined on $(0,\infty)$ and $(-\infty,0)$, since (7.4.6) can be rewritten as

\[ ay''+{b\over x}y'+{c\over x^2}y=0. \]

For convenience we’ll restrict our attention to the interval $(0,\infty)$. (Exercise 19 deals with solutions of (7.4.6) on $(-\infty,0)$.) The key to finding solutions on $(0,\infty)$ is that if $x>0$ then $x^r$ is defined as a real-valued function on $(0,\infty)$ for all values of $r$, and substituting $y=x^r$ into (7.4.6) produces

\begin{equation} \begin{array}{lcl} ax^2(x^r)''+bx(x^r)'+cx^r&=&ax^2r(r-1)x^{r-2}+bxrx^{r-1}+cx^r\\ &=&[ar(r-1)+br+c]x^r. \end{array} \tag{7.4.7}\end{equation}

The polynomial

\[ p(r)=ar(r-1)+br+c \]

is called the indicial polynomial of (7.4.6), and $p(r)=0$ is its indicial equation. From (7.4.7) we can see that $y=x^r$ is a solution of (7.4.6) on $(0,\infty)$ if and only if $p(r)=0$. Therefore, if the indicial equation has distinct real roots $r_1$ and $r_2$ then $y_1=x^{r_1}$ and $y_2=x^{r_2}$ form a fundamental set of solutions of (7.4.6) on $(0,\infty)$, since $y_2/y_1=x^{r_2-r_1}$ is nonconstant. In this case

\[ y=c_1x^{r_1}+c_2x^{r_2} \]

is the general solution of (7.4.6) on $(0,\infty)$.

Example 7.4.4

Find the general solution of

\begin{equation} x^2y''-xy'-8y=0 \tag{7.4.8}\end{equation}

on $(0,\infty)$.

Solution The indicial polynomial of (7.4.8) is

\[ p(r)=r(r-1)-r-8=(r-4)(r+2). \]

Therefore $y_1=x^4$ and $y_2=x^{-2}$ are solutions of (7.4.8) on $(0,\infty)$, and its general solution on $(0,\infty)$ is

\[ y=c_1x^4+{c_2\over x^2}. \]

Example 7.4.5

Find the general solution of

\begin{equation} 6x^2y''+5xy'-y=0 \tag{7.4.9}\end{equation}

on $(0,\infty)$.

Solution The indicial polynomial of (7.4.9) is

\[ p(r)=6r(r-1)+5r-1=(2r-1)(3r+1). \]

Therefore the general solution of (7.4.9) on $(0,\infty)$ is

\[ y=c_1x^{1/2}+c_2x^{-1/3}. \bbox \]

If the indicial equation has a repeated root $r_1$, then $y_1=x^{r_1}$ is a solution of

\begin{equation} ax^2y''+bxy'+cy=0, \tag{7.4.10}\end{equation}

on $(0,\infty)$, but (7.4.10) has no other solution of the form $y=x^r$. If the indicial equation has complex conjugate zeros then (7.4.10) has no real–valued solutions of the form $y=x^r$. Fortunately we can use the results of Section 5.2 for constant coefficient equations to solve (7.4.10) in any case.

Theorem 7.4.3

Suppose the roots of the indicial equation

\begin{equation} ar(r-1)+br+c=0 \tag{7.4.11}\end{equation}

are $r_1$ and $r_2$. Then the general solution of the Euler equation

\begin{equation} ax^2y''+bxy'+cy=0 \tag{7.4.12}\end{equation}

on $(0,\infty)$ is

\begin{eqnarray*} y&=&c_1x^{r_1}+c_2x^{r_2}\mbox{ if $r_1$ and $r_2$ are distinct real numbers }; \\ y&=&x^{r_1}(c_1+c_2\ln x)\mbox{ if $r_1=r_2$ }; \\ y&=&x^{\lambda}\left[c_1\cos\left(\omega\ln x\right)+ c_2\sin\left(\omega\ln x \right)\right]\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega>0$}. \end{eqnarray*}

Proof We first show that $y=y(x)$ satisfies (7.4.12) on $(0,\infty)$ if and only if $Y(t)=y(e^t)$ satisfies the constant coefficient equation

\begin{equation} a{d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0 \tag{7.4.13}\end{equation}

on $(-\infty,\infty)$. To do this, it’s convenient to write $x=e^t$, or, equivalently, $t=\ln x$; thus, $Y(t)=y(x)$, where $x=e^t$. From the chain rule,

\[ {dY\over dt}={dy\over dx}{dx\over dt} \]

and, since

\[ {dx\over dt}=e^t=x, \]

it follows that

\begin{equation} {dY\over dt}=x{dy\over dx}. \tag{7.4.14}\end{equation}

Differentiating this with respect to $t$ and using the chain rule again yields

\begin{eqnarray*} {d^2Y\over dt^2}&=&{d\phantom{t}\over dt}\left(dY\over dt\right)={d\phantom{t}\over dt}\left(x{dy\over dx}\right) \\[6pt] &=&{dx\over dt}{dy\over dx}+x{d^2y\over dx^2}{dx\over dt} \\[6pt] &=&x{dy\over dx}+x^2{d^2y\over dx^2}\quad\left(\mbox{ since } {dx\over dt}=x\right). \end{eqnarray*}

From this and (7.4.14),

\[ x^2{d^2y\over dx^2}={d^2Y\over dt^2}-{dY\over dt}. \]

Substituting this and (7.4.14) into (7.4.12) yields (7.4.13). Since (7.4.11) is the characteristic equation of (7.4.13), Theorem 5.2.1 implies that the general solution of (7.4.13) on $(-\infty,\infty)$ is

\begin{eqnarray*} Y(t)&=&c_1e^{r_1t}+c_2e^{r_2t}\mbox{ if $r_1$ and $r_2$ are distinct real numbers; } \\ Y(t)&=&e^{r_1t}(c_1+c_2t)\mbox{ if $r_1=r_2$; } \\ Y(t)&=&e^{\lambda t }\left(c_1\cos\omega t+c_2\sin\omega t \right)\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega\ne0$}. \end{eqnarray*}

Since $Y(t)=y(e^t)$, substituting $t=\ln x$ in the last three equations shows that the general solution of (7.4.12) on $(0,\infty)$ has the form stated in the theorem.

Example 7.4.6

Find the general solution of

\begin{equation} x^2y''-5xy'+9y=0 \tag{7.4.15}\end{equation}

on $(0,\infty)$.

Solution The indicial polynomial of (7.4.15) is

\[ p(r)=r(r-1)-5r+9=(r-3)^2. \]

Therefore the general solution of (7.4.15) on $(0,\infty)$ is

\[ y=x^3(c_1+c_2 \ln x). \]

Example 7.4.7

Find the general solution of

\begin{equation} x^2y''+3xy'+2y=0 \tag{7.4.16}\end{equation}

on $(0,\infty)$.

Solution The indicial polynomial of (7.4.16) is

\[ p(r)=r(r-1)+3r+2=(r+1)^2+1. \]

The roots of the indicial equation are $r=-1 \pm i$ and the general solution of (7.4.16) on $(0,\infty)$ is

\[ y={1\over x}\left[c_1\cos(\ln x)+c_2\sin(\ln x)\right]. \]

7.4 Exercises

In Exercises 1–18 find the general solution of the given Euler equation on $(0,\infty)$.

$x^2y''+7xy'+8y=0$

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$y=c_1x^{-4}+c_2x^{-2}$
$x^2y''-7xy'+7y=0$

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$y=c_1x+c_2x^7$
$x^2y''-xy'+y=0$

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$y=x(c_1+c_2 \ln x)$
$x^2y''+5xy'+4y=0$

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$y=x^{-2}(c_1+c_2 \ln x)$
$x^2y''+xy'+y=0$

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$y=c_1 \cos (\ln x)+c_2 \sin (\ln x)$
$x^2y''-3xy'+13y=0$

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$y=x^2[c_1 \cos (3 \ln x)+c_2 \sin (3 \ln x)]$
$x^2y''+3xy'-3y=0$

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$y=\dst{c_1x+{c_2\over x^3}}$
$12x^2y''-5xy''+6y=0$

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$y=c_1x^{2/3}+c_2 x^{3/4}$
$4x^2y''+8xy'+y=0$

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$y=x^{-1/2} (c_1+c_2 \ln x)$
$3x^2y''-xy'+y=0$

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$y=c_1x+c_2x^{1/3}$
$2x^2y''-3xy'+2y=0$

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$y=c_1x^2+c_2 x^{1/2}$
$x^2y''+3xy'+5y=0$

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$y=\dst{ {1\over x}\left[c_1\cos(2 \ln x)+c_2\sin(2 \ln x\right]}$
$9x^2y''+15xy'+y=0$

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$y=x^{-1/3} (c_1+c_2 \ln x)$
$x^2y''-xy'+10y=0$

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$y=x\left[c_1\cos(3 \ln x)+c_2\sin(3 \ln x)\right]$
$x^2y''-6y=0$

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$y=\dst{c_1x^3+{c_2\over x^2}}$
$2x^2y''+3xy'-y=0$

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$y=\dst{{c_1\over x}+c_2 x^{1/2}}$
$x^2y''-3xy'+4y=0$

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$y=x^2(c_1+c_2 \ln x)$
$2x^2y''+10xy'+9y=0$

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$y=\dst{{1\over x^2} \left[c_1\cos\left({1\over\sqrt{2}} \ln x\right)+c_2\sin \left({1\over\sqrt{2}} \ln x\right) \right]}$
1. Adapt the proof of Theorem 7.4.3 to show that $y=y(x)$ satisfies the Euler equation
  
  \begin{equation} ax^2y''+bxy'+cy=0 \tag{7.4.1}\end{equation}
  
  on $(-\infty,0)$ if and only if $Y(t)=y(-e^t)$
  
  \[ a {d^2Y\over dt^2}+(b-a){dY\over dt}+cY=0. \]
  
  on $(-\infty,\infty)$.
2. Use (a) to show that the general solution of (7.4.1) on $(-\infty,0)$ is
  
  \begin{eqnarray*} y&=&c_1|x|^{r_1}+c_2|x|^{r_2}\mbox{ if $r_1$ and $r_2$ are distinct real numbers; } \\ y&=&|x|^{r_1}(c_1+c_2\ln|x|)\mbox{ if $r_1=r_2$; } \\ y&=&|x|^{\lambda}\left[c_1\cos\left(\omega\ln|x|\right)+ c_2\sin\left(\omega\ln|x| \right)\right]\mbox{ if $r_1,r_2=\lambda\pm i\omega$ with $\omega>0$}. \end{eqnarray*}
Use reduction of order to show that if

\[ ar(r-1)+br+c=0 \]

has a repeated root $r_1$ then $y=x^{r_1}(c_1+c_2\ln x)$ is the general solution of

\[ ax^2y''+bxy'+cy=0 \]

on $(0,\infty)$.
A nontrivial solution of

\[ P_0(x)y''+P_1(x)y'+P_2(x)y=0 \]

is said to be oscillatory on an interval $(a,b)$ if it has infinitely many zeros on $(a,b)$. Otherwise $y$ is said to be nonoscillatory on $(a,b)$. Show that the equation

\[ x^2y''+ky=0 \quad (k=\; \mbox{constant}) \]

has oscillatory solutions on $(0,\infty)$ if and only if $k>1/4$.
In Example 7.4.2 we saw that $x_0=1$ and $x_0=-1$ are regular singular points of Legendre’s equation

\[ (1-x^2)y''-2xy'+\alpha(\alpha+1)y=0. \tag*{\rm (A)} \]
1. Introduce the new variables $t=x-1$ and $Y(t)=y(t+1)$, and show that $y$ is a solution of (A) if and only if $Y$ is a solution of
  
  \[ t(2+t){d^2Y\over dt^2}+2(1+t){dY\over dt}-\alpha(\alpha+1)Y=0, \]
  
  which has a regular singular point at $t_0=0$.
2. Introduce the new variables $t=x+1$ and $Y(t)=y(t-1)$, and show that $y$ is a solution of (A) if and only if $Y$ is a solution of
  
  \[ t(2-t){d^2Y\over dt^2}+2(1-t){dY\over dt}+\alpha(\alpha+1)Y=0, \]
  
  which has a regular singular point at $t_0=0$.
Let $P_0,P_1$, and $P_2$ be polynomials with no common factor, and suppose $x_0\ne0$ is a singular point of

\[ P_0(x)y''+P_1(x)y'+P_2(x)y=0. \tag*{\rm (A)} \]

Let $t=x-x_0$ and $Y(t)=y(t+x_0)$.
1. Show that $y$ is a solution of (A) if and only if $Y$ is a solution of
  
  \[ R_0(t){d^2Y\over dt^2}+R_1(t){dY\over dt}+R_2(t)Y=0. \tag*{\rm (B)} \]
  
  where
  
  \[ R_i(t)=P_i(t+x_0),\quad i=0,1,2. \]
2. Show that $R_0$, $R_1$, and $R_2$ are polynomials in $t$ with no common factors, and $R_0(0)=0$; thus, $t_0=0$ is a singular point of (B).